Question

Polonium has a half-life of $140$ days. If we take $20\,g$ of polonium initially then the amount of it that remains after $280$ days is

• A. 2.5 g
• B. 5 g
• C. 10 g
• D. 15 g

Answer 1

Answer: (B)

5 g

Answer 2

After n-half lines, quantity of radiatioactive substance left is
$N=N_{0}\left(\frac{1}{2}\right)^{t / T_{1 / 2}}$
Given, $T_{1 / 2}=140$ days,
$t=28$ days, $N_{0}=20\, g$
$\therefore N=20\left(\frac{1}{2}\right)^{\frac{280}{140}}$
$=\frac{20}{4}=5\, g$